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Perturbation question

  1. Jun 5, 2007 #1
    The higher order psi are written with larger epsilon factors in front but the higher order psi are precisely the ones that are meant to more exactly approach the purturbed system however we are decreasing its importance. Is the reason because they can also more easily deviate greater from the true solution?

    It seems that in any purturbed system we are increasing the energy? Or can we decrease it as well? But if we decrease it we will still be adding the higher order energies which may be less then the unperturbed system but adding positive terms will always mean increasing in energy.

  2. jcsd
  3. Jun 5, 2007 #2
    Perturbations can both increase or decrease the total energy of the system. Just look at first order corrections:

    \Delta E^{(1)}_n = \langle n|V|n\rangle

    Why would you assume that this matrix element has to be positive?

    Also, your perturbative expansion is really a Taylor expansion in powers of a "small parameter", so that if your hamiltonian looks like

    \mathcal{H} = \mathcal{H}_0 + \epsilon V

    you assume that the energy eigenstates and eigenvalues are functions of this small parameter:

    [tex] \mathcal{H} |\Psi(\epsilon)\rangle = E(\epsilon) |\Psi(\epsilon)\rangle[/tex]

    From here, you can look at each order of perturbation theory as calculating the expansion coefficients for the energy and the state vector.
    Last edited: Jun 5, 2007
  4. Jun 5, 2007 #3
    The higher-order terms do more closely approach the true solution, but they are also the last important terms, with the least effect on the result. The situation is just like that of a Taylor expansion. Suppose I want to calculate sin(eps) where eps is small. Then I get a perturbation series

    eps - eps^3/6 + eps^5/120 - ...

    Including higher- and higher-order terms get me closer & closer to the solution, but they become progressively smaller and less important as I go on. I can get a very good approximation just by considering the first term or two. That's the magic of perturbation theory: it allows us to take very complicated interactions and calculate out approximate results with just a few steps.


    Like SMG said there's nothing about V that makes it have to be positive. First, remember that it's an operator, not a c-number. Second, even just considering traditional potentials of the form V(x), there's of course many purely negative potentials... the Coulomb or gravitational potentials for example.
  5. Jun 5, 2007 #4
    [tex]\E^{(1)}_n[/tex] is the increase in energy due to the perturbed system as a first approximation.

    What is the meaning of [tex]\E^{(2)}_n[/tex]? Is it the increase in energy due to the perturbed system as a second approximation? Or something else since as damgo said it does not mean a closer approximation.
  6. Jun 5, 2007 #5
    ^^^ It does give you a better approximation. Here, maybe this will make it clear. The exact energy of the perturbed system is

    n0 + n1 + n2 + n3 + n4 + ....

    where n0 is the energy of the unperturbed system, and n1, n2, etc are the corrections you're talking about. So we need to add them all to the original unperturbed energy to get the exact answer.

    But, they have the nice property***

    n1 > n2 > n3 > n4 > ....

    So the corrections keep shrinking in size; otherwise we couldn't do perturbation theory! In fact they tend to decrease quite rapidly -- eg n2+n3+n4+... is usually smaller than n1, and so on. Therefore we can get a good result by neglecting the later terms, since they are so small compared with the early ones.

    So n0+n1 is our "first-order" approx to the true energy, n0+n1+n2 is the better "second-order" approx to the true energy, n0+n1+n2+n3 is an even better approximation, and so on. In practice, you almost never bother going past n0+n1+n2 -- usually that's gives you a very good answer.

    *** Actually this isn't necessarily true... you can have situations in which the n^(i)'s increase at first, as long as they start decreasing sufficiently rapidly at some point.
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